1020. Tree Traversals (25) @ PAT (Advanced Level) Practise

思路：

对后序遍历结果从后向前进行遍历，其中每个值都是一个子树的根节点。用findNode（我用前序遍历）在tree中找到这个Node，将这个Node内的向量vct（中序遍历信息）拆分成左和右子树。

例：

Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

#include<iostream>#include<vector>using namespace std;struct Node{int value;vector<int> vct;//Node * lNode;//leftNodeNode * rNode;//rightNodeNode(int _value):value(_value),lNode(NULL),rNode(NULL){}//no useNode():value(0),lNode(NULL),rNode(NULL){}};vector<int> PO;//postOrderstruct Node tree;//root nodeNode * findNode(int n,Node * s)//post Traversal the tree and scan the vector vct of the node//n - looking for;s -  start node{Node * np;//node pointerif((*s).value!=0){if(s->rNode!=NULL){if((np=findNode(n,(*s).rNode))!=NULL){return np;}}if(s->lNode!=NULL){if((np=findNode(n,(*s).lNode))!=NULL){return np;}}return NULL;}else{vector<int>::iterator it;for(it=s->vct.begin();it!=s->vct.end();it++){if((*it)==n){return s;}}return NULL;}return np;}int main(){//initialize & read dataint N;cin>>N;PO.resize(N);for(int i=0;i<N;i++)cin>>PO[i];int tmp;for(int i=0;i<N;i++){cin>>tmp;tree.vct.push_back(tmp);}tree.value=0;//generate the treevector<int>::reverse_iterator rit;Node * np;for(rit=PO.rbegin();rit!=PO.rend();rit++){np=findNode(*rit,&tree);(*np).value=*rit;bool pushToLeft=true;vector<int> ::iterator it;for(it=(*np).vct.begin();it!=(*np).vct.end();it++){if(*it == *rit){pushToLeft=false;continue;}else{if(pushToLeft){if((*np).lNode==NULL)(*np).lNode=new Node();(*np).lNode->vct.push_back(*it);}else{if((*np).rNode==NULL)(*np).rNode=new Node();(*np).rNode->vct.push_back(*it);}}}}//output level ordervector<Node *> v1,v2;vector<int > result;v1.push_back(&tree);while(!v1.empty()){for(int i=0;i<v1.size();i++){result.push_back(v1[i]->value);if(v1[i]->lNode!=NULL)v2.push_back(v1[i]->lNode);if(v1[i]->rNode!=NULL)v2.push_back(v1[i]->rNode);}v1.clear();v1.assign(v2.begin(),v2.end());v2.clear();}for(int i=0;i<result.size()-1;i++)cout<<result[i]<<" ";cout<<result[result.size()-1];return 0;}